Answer
Verified
432.6k+ views
Hint: In this problem we have given a function and asked to calculate the integration value. We can observe that the given function is the multiplication of the trigonometric function with the hyperbolic function. First of all, the given function has multiplication operation, so we are going to use the integration by parts rule which is $\int{uvdx}=u\int{vdx}-\int{\left[ \left( {{u}^{'}} \right)\int{vdx} \right]dx}$. So, we will choose $u$, $v$ by ILATE rule. After choosing $u$, $v$. We will apply the integration by parts rule and simplify the equation by using the integration and differentiation formulas.
Complete step by step answer:
Given that, $\left( \cos x \right)\left( \cosh x \right)dx$.
In the above function we have trigonometric function $\cos x$, hyperbolic function $\cosh x$. From ILATE (Inverse, Logarithmic, Algebraic, Trigonometric, Exponential) rule the first function $u$ is $\cos x$, the second function $v$ is $\cosh x$.
Now the integration of the given function from the integration by parts formula $\int{uvdx}=u\int{vdx}-\int{\left[ \left( {{u}^{'}} \right)\int{vdx} \right]dx}$ is given by
$\int{\cos x\cosh xdx}=\cos x\int{\cosh xdx}-\int{\left[ {{\left( \cos x \right)}^{'}}\int{\cosh xdx} \right]dx}$
We have the differentiation formula $\dfrac{d}{dx}\left( \cos x \right)=-\sin x$, applying this formula in the above equation, then we will get
$\Rightarrow \int{\cos x\cosh xdx}=\cos x\int{\cosh xdx}-\int{\left[ \left( -\sin x \right)\int{\cosh xdx} \right]dx}$
We have the integration formula $\int{\cosh xdx}=\sinh x+C$, applying this formula in the above equation, then we will have
$\Rightarrow \int{\cos x\cosh xdx}=\cos x\left( \sinh x \right)+\int{\sin x\sinh xdx}...\left( \text{i} \right)$
From the above equation we can say that to calculate the integration of the given function we need to have the value of $\int{\sin x\sinh xdx}$. So, applying the same integration by parts formula for this function also, then we will have
$\Rightarrow \int{\sin x\sinh xdx}=\sin x\int{\sinh xdx}-\int{\left[ {{\left( \sin x \right)}^{'}}\int{\sinh xdx} \right]dx}$
We have the formulas $\dfrac{d}{dx}\left( \sin x \right)=\cos x$, $\int{\sinh xdx}=\cosh x+C$. Substituting these formulas in the above equation, then we will get
$\Rightarrow \int{\sin x\sinh xdx}=\sin x\cosh x-\int{\cos x\cosh xdx}...\left( \text{ii} \right)$
Substituting this value in the equation $\left( \text{i} \right)$, then we will get
$\Rightarrow \int{\cos x\cosh xdx}=\cos x\sinh x+\left[ \sin x\cosh x-\int{\cos x\cosh xdx} \right]$
Simplifying the above equation, then we will have
$\begin{align}
& \Rightarrow \int{\cos x\cosh xdx}+\int{\cos x\cosh xdx}=\cos x\sinh x+\sin x\cosh x \\
& \Rightarrow 2\int{\cos x\cosh xdx}=\cos x\sinh x+\sin x\cosh x \\
& \therefore \int{\cos x\cosh xdx}=\dfrac{\cos x\sinh x+\sin x\cosh x}{2}+C \\
\end{align}$
Hence the integration of the given function $\left( \cos x \right)\left( \cosh x \right)dx$ is $\dfrac{\cos x\sinh x+\sin x\cosh x}{2}+C$.
Note: In this problem students may make a little mistake which will cost the whole problem. In integration of trigonometric ratios, we have $\int{\sin xdx}=-\cos x+C$, $\int{\cos xdx}=\sin x+C$ but when it comes to integration of hyperbolic functions, we have $\int{\sinh xdx}=\cosh x+C$, $\int{\cosh xdx}=\sinh x+C$. If you forget about this change and solved the problem you will get an incorrect solution.
Complete step by step answer:
Given that, $\left( \cos x \right)\left( \cosh x \right)dx$.
In the above function we have trigonometric function $\cos x$, hyperbolic function $\cosh x$. From ILATE (Inverse, Logarithmic, Algebraic, Trigonometric, Exponential) rule the first function $u$ is $\cos x$, the second function $v$ is $\cosh x$.
Now the integration of the given function from the integration by parts formula $\int{uvdx}=u\int{vdx}-\int{\left[ \left( {{u}^{'}} \right)\int{vdx} \right]dx}$ is given by
$\int{\cos x\cosh xdx}=\cos x\int{\cosh xdx}-\int{\left[ {{\left( \cos x \right)}^{'}}\int{\cosh xdx} \right]dx}$
We have the differentiation formula $\dfrac{d}{dx}\left( \cos x \right)=-\sin x$, applying this formula in the above equation, then we will get
$\Rightarrow \int{\cos x\cosh xdx}=\cos x\int{\cosh xdx}-\int{\left[ \left( -\sin x \right)\int{\cosh xdx} \right]dx}$
We have the integration formula $\int{\cosh xdx}=\sinh x+C$, applying this formula in the above equation, then we will have
$\Rightarrow \int{\cos x\cosh xdx}=\cos x\left( \sinh x \right)+\int{\sin x\sinh xdx}...\left( \text{i} \right)$
From the above equation we can say that to calculate the integration of the given function we need to have the value of $\int{\sin x\sinh xdx}$. So, applying the same integration by parts formula for this function also, then we will have
$\Rightarrow \int{\sin x\sinh xdx}=\sin x\int{\sinh xdx}-\int{\left[ {{\left( \sin x \right)}^{'}}\int{\sinh xdx} \right]dx}$
We have the formulas $\dfrac{d}{dx}\left( \sin x \right)=\cos x$, $\int{\sinh xdx}=\cosh x+C$. Substituting these formulas in the above equation, then we will get
$\Rightarrow \int{\sin x\sinh xdx}=\sin x\cosh x-\int{\cos x\cosh xdx}...\left( \text{ii} \right)$
Substituting this value in the equation $\left( \text{i} \right)$, then we will get
$\Rightarrow \int{\cos x\cosh xdx}=\cos x\sinh x+\left[ \sin x\cosh x-\int{\cos x\cosh xdx} \right]$
Simplifying the above equation, then we will have
$\begin{align}
& \Rightarrow \int{\cos x\cosh xdx}+\int{\cos x\cosh xdx}=\cos x\sinh x+\sin x\cosh x \\
& \Rightarrow 2\int{\cos x\cosh xdx}=\cos x\sinh x+\sin x\cosh x \\
& \therefore \int{\cos x\cosh xdx}=\dfrac{\cos x\sinh x+\sin x\cosh x}{2}+C \\
\end{align}$
Hence the integration of the given function $\left( \cos x \right)\left( \cosh x \right)dx$ is $\dfrac{\cos x\sinh x+\sin x\cosh x}{2}+C$.
Note: In this problem students may make a little mistake which will cost the whole problem. In integration of trigonometric ratios, we have $\int{\sin xdx}=-\cos x+C$, $\int{\cos xdx}=\sin x+C$ but when it comes to integration of hyperbolic functions, we have $\int{\sinh xdx}=\cosh x+C$, $\int{\cosh xdx}=\sinh x+C$. If you forget about this change and solved the problem you will get an incorrect solution.
Recently Updated Pages
Fill in the blanks with suitable prepositions Break class 10 english CBSE
Fill in the blanks with suitable articles Tribune is class 10 english CBSE
Rearrange the following words and phrases to form a class 10 english CBSE
Select the opposite of the given word Permit aGive class 10 english CBSE
Fill in the blank with the most appropriate option class 10 english CBSE
Some places have oneline notices Which option is a class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How do you graph the function fx 4x class 9 maths CBSE
Which are the Top 10 Largest Countries of the World?
What is the definite integral of zero a constant b class 12 maths CBSE
Distinguish between the following Ferrous and nonferrous class 9 social science CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE